# Two unknown charges accelerating an object

## Homework Statement

Two point charges q1 and q2 are held in place 4.50 cm apart. Another point charge -2.00μC of mass 6.00g is initially located 3.00 cm from each of these charges and released from rest. You observe that the initial acceleration of -2.00μC is 354m/s^2 upward, parallel to the line connecting the two point charges. Find q1 and q2.

## Homework Equations

## F=ma ##
## F= kQq1/r^2 ##

## The Attempt at a Solution

I drew a sketch of the problem, with q1 directly above q2 (.045m away). The first thing I did was find the total force by plugging in the mass and acceleration:

## F = (0.006kg)*(354m/s^2) = 2.124 N ##

Next, I saw that since the object is accelerating upward and parallel to the line connecting the two point charges, then q1 must be positive and q2 must be negative. Since Q is moving straight up, there is no movement in the x direction.

With that, I figured that:

## F = Fy ##
## 2.124 = 2 FQy ##
## FQy = 1.062 ##

So far, I get that the total force is the vector sum of Fq1y and Fq2y. And since these vectors are in the y-direction, you can add their magnitudes (I think). I tried solving for q1 in Coulombs' law:

## F*r^2/kQ = q1 ##
## (1.062)(.03^2)/(8.99*10^9)(2*10^-6) = 5.3*10^-8 C ##

(That should be 10^-6 but latex keeps putting the - sign up and not the 6. Not sure how to fix that but I calculated it correctly at that part)

But this isn't the right answer. I'm pretty sure that q1 and q2 have the same magnitude, just different charges (q1 is + and q2 is -), but I don't know what to do in order to find them. Any help would be appreciated. Thanks!

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TSny
Homework Helper
Gold Member
Hello, Ozmahn. Welcome to PF!

You stated correctly that you need to work with the y-components of the electric forces on Q. But in your calculations, I don't see where you actually worked with the y-component of either electric force on Q. Note that the given distances imply that Q cannot be located on the line connecting q1 and q2.

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Thanks TSny!

So I assumed that since Q is accelerating parallel to the line connecting q1 and q2, the sum of the y-components (Q1y and Q2y) is equivalent to the net force. I think the y-components of the vector from q1 to Q and q2 to Q are both equivalent in magnitude, just the charges themselves are opposite signs. And since they're y-components, their vectors are equivalent as well since q1 and q2 are at an equal distance from Q. So

## Fy=FQ1y+FQ2y=2FQy ##
## 2.124=2FQy ##
## 1.062=FQy ##

And that 1.062 is the magnitude of the y-component of the vector created by q1 and q2. And since the y-component is the net force, then plugging that in to Coulomb's law should solve it...at least, that's how I'm thinking about it but something is off. Am I looking at this the wrong way?

TSny
Homework Helper
Gold Member
## Fy=FQ1y+FQ2y=2FQy ##
## 2.124=2FQy ##
## 1.062=FQy ##
This looks good. However, be careful when you proceed to find q1. Keep in mind that you are working with the y-component of the force.

Ah okay, so just because both x-components cancel each other out, they still contribute to the force of their respective q's. So...I need to find the actual force of q1 and q2, not just its y-component? And I'm pretty sure I can do that with arcsin, and since the forces are equal in magnitude but have opposite charges. I'll try it out when I get home from class, thanks!

TSny
Homework Helper
Gold Member
Just to make sure we're together. How would you write the expression for the magnitude of the force of q1 on Q? How would you then write the y-component of this force?

Okay so here is what I did. I already knew what Q1y was (1.062), so I found the angle created by q1x and q1. To do this:

## tan\theta\\ = 0.0225/0.03 => arctan(0.0225/0.03) = 37 ##

So now that I know what the angle is, I can solve the smaller triangle created by q1, q1x, and q1y.

## sin\theta\ = q1y/q1 => q1 = q1y/sin\theta\ => 1.062/0.6 = 1.77 = q1 ##

Now that I know what force q1 is putting on Q, and I've already identified that q1 must be positive and q2 must be negative, I know their absolute value is 1.77 N. Now to find q1, I can use Coulomb's law:

## F = (k*Q*q1)/r^2 ##

Rewritten to solve for q1:

## q1= (F*r^2)/(k*Q) ##
## q1= (1.77*0.03^2)/(8.99*10^9*2) ##
## q1=8.86*10^-14 ##

And q2 would be the same, but negative. How does that sound?

TSny
Homework Helper
Gold Member
Okay so here is what I did. I already knew what Q1y was (1.062), so I found the angle created by q1x and q1. To do this:

## tan\theta\\ = 0.0225/0.03 => arctan(0.0225/0.03) = 37 ##
I don't believe the tangent function is correct here. See diagram below.

So now that I know what the angle is, I can solve the smaller triangle created by q1, q1x, and q1y.

## sin\theta\ = q1y/q1 => q1 = q1y/sin\theta\ => 1.062/0.6 = 1.77 = q1 ##
Your notation is kind of dangerous. By q1 and q1y here, I guess you mean the forces Fq1Q and Fq1Q,y.

Now that I know what force q1 is putting on Q, and I've already identified that q1 must be positive and q2 must be negative, I know their absolute value is 1.77 N. Now to find q1, I can use Coulomb's law:

## F = (k*Q*q1)/r^2 ##

Rewritten to solve for q1:

## q1= (F*r^2)/(k*Q) ##
## q1= (1.77*0.03^2)/(8.99*10^9*2) ##
## q1=8.86*10^-14 ##

And q2 would be the same, but negative. How does that sound?
OK, but Q is not 2 Coulombs.

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I attached a picture of the system, this was provided by the website. I should have included that with the initial post, I didn't think it could be interpreted in the other direction.

Sorry about the notation, I should definitely be more clear about that. Yes, that is what I meant. I just realized how to format subscripts, so I'll start doing that with future posts.

Finally, it should be 2x10-6 since it's in μC.

I think I got it this time...

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Ohh I see what you're saying. It shouldn't be tangent at all, I should be using opposite over hypotenuse. So sine instead.

The FQq1 comes out to 1.42 N, so plugging that in to solve for q1 I get 7.11x10-8.

That was the correct answer. Thanks a lot for the help, I appreciate it!

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TSny
Homework Helper
Gold Member
Ok. Good work!